Explanation​:

Induced emf is proportional with rate of change in magnetic flux. As the distance between the bar magnet and loop magnetic flux changes and the rate of change increases as the relative velocity increases (vL>vK=vM​).

Therefore, εL>εK=εM​

According to Fleming’s left hand rule, the electrons in the rod experience a force towards Side B at the top of the rod. The flow of electrons (negative charges) towards the top end of the rod will stop when the electrons at that end are numerous enough to push new electrons back by electrostatic repulsion. That is, an electric field is established in the rod where the top is negative and the bottom is positive. The corresponding field is from the bottom to the top (Side D to Side B)

Explanation​:
The alpha particle has positive charge, therefore its deflection will be in the opposite direction of the electron. (answer A or B)

It has a larger mass, therefore the acceleration will be much smaller, explained by the following calculation:

• qE=ma
• a=qE/m​

The alpha particle has twice the charge of the electron, this doubles its acceleration.

It has much more mass than the electron, therefore the denominator in the above ratio increases by a larger factor than the numerator. This means the acceleration of the alpha particle will be much smaller than that of the electron. (its motion will change at a slower rate).

This can also be explained by 2-dimensional motion where there is no force on the horizontal but there is a smaller force on the vertical, meaning it will cover the vertical distance in a longer time, allowing it to travel further along the horizontal, giving the path in B.

Explanation​:

When the bar magnet moves towards and away from the ring, the magnetic flux through the ring will change. According to Lenz's law, an emf, and therefore a current is induced to oppose the change in the flux. So as the bar magnet approaches the metal ring, the metal ring moves to the right due to repulsive forces that resist the decrease in distance between the ring and the magnet. Likewise, as the bar magnet moves away from the metal ring, the metal ring moves to the left.

Explanation​:
Recall that magnetic force acts perpendicular to current flow according to the Fleming Left hand rule. That immediately eliminates option A and D. Then using the right hand grip rule, we can determine that the magnetic field experienced by wire $Y$ points into the page. The force direction and magnetic field direction allows current direction to be found with Fleming’s left-hand rule.

Explanation​:

• I. Correct: Since the particle is deflected toward to negative plate it is positively charged.
• II. Correct: In order not to be deflected magnetic force acts on the particle should be upwards. According to the right hand rule the magnetic field should be into the page.
• III. Incorrect: In order not to be deflected magnetic force acts on the particle should be in upwards. According to the right hand rule magnetic field should be into the page.

Explanation​:
The gravitational field strength varies inversely with the square of the distance from the center of the planet ($R$) to a given point.

g=GM(R)^2​

This means the variation of the gravitational field strength with $R$ is an inverse square law. Note, however, that the question asks for the relationship from the distance from the surface of the planet, therefore the graph starts at a finite value of field strength at the surface ($r=0$), then follows the shape of an inverse square curve.

Explanation​:

Using the right-hand rule, one can determine the direction of magnetic fields on loops $Y$ and $Z$ due to the current in the wire.

The magnetic field on loop $Y$ is in the direction out of the page, and the magnetic field on loop $Z$ is in the direction into the page. Both decrease as the current in the wire decreases, so magnetic flux decreases for both loops $Y$ and $Z$.

According to Lenz' law, a current is induced on the loop $Y$ in an anti-clockwise direction, so it opposes the change in the flux. Likewise, according to Lenz' law, a current is induced on the loop Z in a clockwise direction, so it opposes the change in the flux.

Explanation​:
The $20nC$ charge produces a leftward electric field in region 1. Therefore, the electric field generated by the $X$ charge should be rightward to cancel the electric field of the 20 nC charge. This is achieved by a negative charge at X. The magnitude of the electric field decreases by distance. All locations in region 1 are closer to point charge $X$ than they are to the $20nC$ charge. Therefore X should be smaller than the charge of $20nC$. The answer is D.

Explanation​:

The hydrogen nucleus is a proton, so it has positive charge.

The electric field lines are pointed upwards, meaning the higher potential is below the nucleus, or meaning that the force on the nucleus is upward.

The question asks for the largest work done, therefore we are looking for the largest displacement parallel and in the opposite direction the field lines. This is shown in D.

1/4

Explanation​:

Because the electric field lines point toward the charge $q$, this charge must be negative, meaning that statement I is false.

The relatively greater number of field lines around charge $q$ means that this charge magnitude is greater than the charge +Q, indicating that statement II is also untrue.

Electric field line density is proportional to electric field strength. Because the field line density near $q$ is greater than that near $+Q$, statement III is correct.

Therefore the correct answer is option C.

4g

Explanation​:
The electric field vector inside a hollow conductor is zero. Due to the lack of field inside the sphere, the electric potential is constant and equal to the surface value . Outward from the surface of the sphere, the electric field varies with the inverse​ square of the distance from the center while electric potential decreases inversely with the distance from the center.

Explanation​:
A is incorrect because the equipotentials have equal potential difference but the lines are not equally spaced.

B is incorrect because work =qΔV=(−e)[−30 V−(−50 V)]=−20 eV

C is correct because work =qΔV=(−e)[−30 V−(−10 V)]=20 eV

D is incorrect because work done from R to S is zero (same potential)

Explanation​:
The force acting on the electron is downward at entry and perpendicular to the velocity, and the electron has negative charge. Using conventional methods (right-hand/left-hand) the magnetic field vector points into the plane of the paper

Explanation​:
At a level closer to the surface of the planet, the centripetal force increases. In a circular orbit, the centripetal force is provided by the gravitational force, which is proportional to the inverse square of the distance from the center of the planet.

The kinetic energy in a closer orbit is larger since the centripetal force is larger. Making the speed larger is necessary to maintain the orbit, which therefore increases the kinetic energy.